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Math Help - infinite series

  1. #1
    Junior Member ihmth's Avatar
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    infinite series

    Find the sum of the infinite series
    1/5 + 2/5^2 + 3/5^3 + 1/5^4 + 2/5^5 + 3/5^6 + 1/5^7 + 2/5^8 + 3/5^9 + .....
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  2. #2
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    Quote Originally Posted by ihmth View Post
    Find the sum of the infinite series
    1/5 + 2/5^2 + 3/5^3 + 1/5^4 + 2/5^5 + 3/5^6 + 1/5^7 + 2/5^8 + 3/5^9 + .....
    Re-arrange:

    = (1/5 + 1/5^4 + 1/5^7 + ....) + (2/5^2 + 2/5^5 + 2/5^8 + .....) + (3/5^3 + 3/5^6 + 3/5^9 + .....)

    = 1/5 (1 + 1/5^3 + 1/5^6 + ....) + 2/5^2 (1 + 1/5^3 + 1/5^6 + ....) + 3/5^3 (1 + 1/5^3 + 1/5^6 + ......)

    = (1/5 + 2/5^2 + 3/5^3) (1 + 1/5^3 + 1/5^6 + ....)

    = 38/5^3 (1 + 1/5^3 + 1/5^6 + ....)

    where (1 + 1/5^3 + 1/5^6 + ....) is an infinite geometric series a = 1 and r = 1/5^3. And you know how to get that sum, right?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ihmth View Post
    Find the sum of the infinite series
    1/5 + 2/5^2 + 3/5^3 + 1/5^4 + 2/5^5 + 3/5^6 + 1/5^7 + 2/5^8 + 3/5^9 + .....
    What Mr Fantastic said, but also if you group the terms in threes you will
    find its a geometric series and we all know how to sum those.

    RonL
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  4. #4
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    Hello, ihmth!

    Yet another approach . . .


    Find the sum of the infinite series:

    . . S \;=\;\frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \frac{1}{5^4} + \frac{2}{5^5} + \frac{3}{5^6} + \frac{1}{5^7} + \frac{2}{5^8} + \frac{3}{5^9} + \hdots

    Multiply both sides by 5^3

    . . 5^3S \;=\;5^2 + 2\!\cdot\!5 + 3 + \underbrace{\frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \frac{1}{5^4} + \frac{2}{5^5} + \frac{3}{5^6} + \hdots}_{\text{This is }S}


    And we have: . 125S \;=\;38 + S\quad\Rightarrow\quad 124S \:=\:38

    Therefore: . \boxed{S \;=\;\frac{19}{62}}

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  5. #5
    Junior Member ihmth's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Re-arrange:

    = (1/5 + 1/5^4 + 1/5^7 + ....) + (2/5^2 + 2/5^5 + 2/5^8 + .....) + (3/5^3 + 3/5^6 + 3/5^9 + .....)

    = 1/5 (1 + 1/5^3 + 1/5^6 + ....) + 2/5^2 (1 + 1/5^3 + 1/5^6 + ....) + 3/5^3 (1 + 1/5^3 + 1/5^6 + ......)

    = (1/5 + 2/5^2 + 3/5^3) (1 + 1/5^3 + 1/5^6 + ....)

    = 38/5^3 (1 + 1/5^3 + 1/5^6 + ....)

    where (1 + 1/5^3 + 1/5^6 + ....) is an infinite geometric series a = 1 and r = 1/5^3. And you know how to get that sum, right?
    Tnx
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